I have read the docs and still not quite following it. According to it, it returns me specific documents according to my own specifications inside a collection. For grouping, it pretty much says the same thing: "Groups documents by some specified expression and outputs to the next stage a document for each distinct grouping"
So, what does this following code is actually doing? It seems redundant to me.
BillingCycle.aggregate([{
$project: {credit: {$sum: "$credits.value"}, debt: {$sum: "debts.value"}}
}, {
$group: {
_id: null,
credit: {$sum: "$credit"}, debt: {$sum: "debt"}
}
}, {
$project: {_id: 0, credit: 1, debt: 1 }
}]});
"Groups documents by some specified expression and outputs to the next stage a document for each distinct grouping"
The purpose of $group is not only to push some fields to next stage but to gather some element on the basis of input criteria passed in the _id attribute.
On the other, hand $project function will exclude/include some field(or custom field) to next stage. As per document you can see the definition "Passes along the documents with the requested fields to the next stage in the pipeline. The specified fields can be existing fields from the input documents or newly computed fields."
There is one case if we suppress the _id from $group then it will calculate accumulated values for all the input documents as a whole. Which seems to act like $project.
For the query on $project stage is redundant
BillingCycle.aggregate([ {
$group: {
_id: null,
credit: {$sum: "$credit.value"}, debt: {$sum: "debt.value"}
}
}, {
$project: {_id: 0, credit: 1, debt: 1 }
}]});
Related
I'm trying to look up all records that match a certain condition, in this case _id being certain values, and then return only the top 2 results, sorted by the name field.
This is what I have
db.getCollection('col1').aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$sort: {fk: 1, name: -1}},
{$group: {_id: "$fk", items: {$push: "$$ROOT"} }},
{$project: {items: {$slice: ["$items", 2]} }}
])
and it works, BUT, it's not guaranteed. According to this Mongo thread $group does not guarantee document order.
This would also mean that all of the suggested solutions here and elsewhere, which recommend using $unwind, followed by $sort, and then $group, would also not work, for the same reason.
What is the best way to accomplish this with Mongo (any version)? I've seen suggestions that this could be accomplished in the $project phase, but I'm not quite sure how.
You are correct in saying that the result of $group is never sorted.
$group does not order its output documents.
Hence doing a;
{$sort: {fk: 1}}
then grouping with
{$group: {_id: "$fk", ... }},
will be a wasted effort.
But there is a silver lining with sorting before $group stage with name: -1. Since you are using $push (not an $addToSet), inserted objects will retain the order they've had in the newly created items array in the $group result. You can see this behaviour here (copy of your pipeline)
The items array will always have;
"items": [
{
..
"name": "Michael"
},
{
..
"name": "George"
}
]
in same order, therefore your nested array sort is a non-issue! Though I am unable to find an exact quote in documentation to confirm this behaviour, you can check;
this,
or this where it is confirmed.
Also, accumulator operator list for $group, where $addToSet has "Order of the array elements is undefined." in its description, whereas the similar operator $push does not, which might be an indirect evidence? :)
Just a simple modification of your pipeline where you move the fk: 1 sort from pre-$group stage to post-$group stage;
db.getCollection('col1').aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$sort: {name: -1}},
{$group: {_id: "$fk", items: {$push: "$$ROOT"} }},
{$sort: {_id: 1}},
{$project: {items: {$slice: ["$items", 2]} }}
])
should be sufficient to have the main result array order fixed as well. Check it on mongoplayground
$group doesn't guarantee the document order but it would keep the grouped documents in the sorted order for each bucket. So in your case even though the documents after $group stage are not sorted by fk but each group (items) would be sorted by name descending. If you would like to keep the documents sorted by fk you could just add the {$sort:{fk:1}} after $group stage
You could also sort by order of values passed in your match query should you need by adding a extra field for each document. Something like
db.getCollection('col1').aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$addField:{ifk:{$indexOfArray:[[1, 2],"$fk"]}}},
{$sort: {ifk: 1, name: -1}},
{$group: {_id: "$ifk", items: {$push: "$$ROOT"}}},
{$sort: {_id : 1}},
{$project: {items: {$slice: ["$items", 2]}}}
])
Update to allow array sort without group operator : I've found the jira which is going to allow sort array.
You could try below $project stage to sort the array.There maybe various way to do it. This should sort names descending.Working but a slower solution.
{"$project":{"items":{"$reduce":{
"input":"$items",
"initialValue":[],
"in":{"$let":{
"vars":{"othis":"$$this","ovalue":"$$value"},
"in":{"$let":{
"vars":{
//return index as 0 when comparing the first value with initial value (empty) or else return the index of value from the accumlator array which is closest and less than the current value.
"index":{"$cond":{
"if":{"$eq":["$$ovalue",[]]},
"then":0,
"else":{"$reduce":{
"input":"$$ovalue",
"initialValue":0,
"in":{"$cond":{
"if":{"$lt":["$$othis.name","$$this.name"]},
"then":{"$add":["$$value",1]},
"else":"$$value"}}}}
}}
},
//insert the current value at the found index
"in":{"$concatArrays":[
{"$slice":["$$ovalue","$$index"]},
["$$othis"],
{"$slice":["$$ovalue",{"$subtract":["$$index",{"$size":"$$ovalue"}]}]}]}
}}}}
}}}}
Simple example with demonstration how each iteration works
db.b.insert({"items":[2,5,4,7,6,3]});
othis ovalue index concat arrays (parts with counts) return value
2 [] 0 [],0 [2] [],0 [2]
5 [2] 0 [],0 [5] [2],-1 [5,2]
4 [5,2] 1 [5],1 [4] [2],-1 [5,4,2]
7 [5,4,2] 0 [],0 [7] [5,4,2],-3 [7,5,4,2]
6 [7,5,4,2] 1 [7],1 [6] [5,4,2],-3 [7,6,5,4,2]
3 [7,6,5,4,2] 4 [7,6,5,4],4 [3] [2],-1 [7,6,5,4,3,2]
Reference - Sorting Array with JavaScript reduce function
There is a bit of a red herring in the question as $group does guarantee that it will be processing incoming documents in order (and that's why you have to sort of them before $group to get an ordered arrays) but there is an issue with the way you propose doing it, since pushing all the documents into a single grouping is (a) inefficient and (b) could potentially exceed maximum document size.
Since you only want top two, for each of the unique fk values, the most efficient way to accomplish it is via a "subquery" using $lookup like this:
db.coll.aggregate([
{$match: {fk: {$in: [1, 2]}}},
{$group:{_id:"$fk"}},
{$sort: {_id: 1}},
{$lookup:{
from:"coll",
as:"items",
let:{fk:"$_id"},
pipeline:[
{$match:{$expr:{$eq:["$fk","$$fk"]}}},
{$sort:{name:-1}},
{$limit:2},
{$project:{_id:0, fk:1, name:1}}
]
}}
])
Assuming you have an index on {fk:1, name:-1} as you must to get efficient sort in your proposed code, the first two stages here will use that index via DISTINCT_SCAN plan which is very efficient, and for each of them, $lookup will use that same index to filter by single value of fk and return results already sorted and limited to first two. This will be the most efficient way to do this at least until https://jira.mongodb.org/browse/SERVER-9377 is implemented by the server.
I want to know how to use aggregate() to take all of the objects of a specific field (i.e. "user") and count them.
This what I am doing:
I want to return a list of users with the sum of how many tweets that have made?
So I want output that looks like
Etc..
Also I don't want repeating users like
Etc..
which is what the above aggregate does.
So basically, how can I modify this aggregate to ensure the objects are unique?
I believe you will want to group by the user.id field instead of the user object. You can try doing that directly
$group: {_id: "$user.id", totalTweets: {$sum: 1} }
Or you might want to try projecting that field onto the document before grouping
$addFields: {userId: "$user.id"}
$group: {_id: "$userId", totalTweets: {$sum: 1} }
If you want whole inner user object in each documents after aggregation then you have to use $push operator in aggregation
and also you need to do the aggregation on unique id of users e.g: id or id_str instead of $user object as in your question.
db.tweets.aggregate([{ $group: {_id: "$user.id", totalTweets: { $sum: 1 }, user : { $push: "$user" } } }])
This will solved your problem. For details about $push operator, have a look at official documents $push
We have a rudimentary versioning system in a collection that uses a field (pageId) as a root key. Subsequent versions of this page have the same pageId. This allows us to very easily find all versions of a single page.
How do I go about running a query that returns only the lastModified document for each distinct pageId.
In psuedo-code you could say:
For each distinct pageId
sort documents based on lastModified descending
and return only the first document
You can use the aggregation pipelines for that.
$sort - Sorts all input documents and returns them to the pipeline in sorted order.
$group - Groups documents by some specified expression and outputs to the next stage a document for each distinct grouping.
$first - Returns the value that results from applying an expression to the first document in a group of documents that share the same group by key.
Example:
db.getCollection('t01').aggregate([
{
$sort: {'lastModified': -1}
},
{
$group: {
_id: "$pageId",
element1: { $first: "$element1" },
element2: { $first: "$element2" },
elementN: { $first: "$elementN" },
}
}
]);
I have the following document:
{
'date': date,
'_id': ObjectId,
'Log': [
{
'lat': float,
'lng': float,
'date': float,
'speed': float,
'heading': float,
'fix': float
}
]
}
for 1 document, the Log array can be some hundred entries.
I need to query the first and last date element of Log on each document. I know how to query it, but I need to do it fast, so I would like to build an index for that. I don't want to index Log.date since it is too big... how can I index them?
In fact it's hard to advise without knowing how you work with the documents. One of the solutions could be to use a sparse index. You just need to add a new field to every first and last array element, let's call it shouldIndex. Then just create a sparse index which includes shouldIndex and date fields. Here's a short example:
Assume we have this document
{"Log":
[{'lat': 1, 'lng': 2, 'date': new Date(), shouldIndex : true},
{'lat': 3, 'lng': 4, 'date': new Date()},
{'lat': 5, 'lng': 6, 'date': new Date()},
{'lat': 7, 'lng': 8, 'date': new Date(), shouldIndex : true}]}
Please note the first element and the last one contain shouldIndex field.
db.testSparseIndex.ensureIndex( { "Log.shouldIndex": 1, "Log.date":1 }, { spar
se: true } )
This index should contain entries only for your first and last elements.
Alternatively you may store first and last elements date field in a seperate array.
For more info on sparse indexes please refer to this article.
Hope it helps!
So there was an answer about indexing that is fundamentally correct. As of writing though it seems a little unclear whether you are talking about indexing at all. It almost seems like what you want to do is get the first and last date from the elements in your array.
With that in mind there are a few approaches:
1. The elements in your array have been naturally inserted in increasing date values
So if the way all writes that are made to this field is done, only with use of the $push operator over a period of time, and you never update these items, at least in so much as changing a date, then your items are already in order.
What this means is you just get the first and last element from the array
db.collection.find({ _id: id },{ Log: {$slice: 1 }}); // gets the first element
db.collection.find({ _id: id },{ Log: {$slice: -1 }}); // gets the last element
Now of course that is two queries but it's a relatively simple operation and not costly.
2. For some reason your elements are not naturally ordered by date
If this is the case, or indeed if you just can't live with the two query form, then you can get the first and last values in aggregation, but using $min and $max modifiers
db.collection.aggregate([
// You might want to match first. Just doing one _id here. (commented)
//{"$match": { "_id": id }},
//Unwind the array
{"$unwind": "$Log" },
//
{"$group": {
"_id": "$_id",
"firstDate": {"$min": "$Log.Date" },
"lastDate": {"$max": "$Log.Date" }
}}
])
So finally, if your use case here is getting the details of the documents that have the first and last date, we can do that as well, mirroring the initial two query form, somewhat. Using $first and $last :
db.collection.aggregate([
// You might want to match first. Just doing one _id here. (commented)
//{"$match": { "_id": id }},
//Unwind the array
{"$unwind": "$Log" },
// Sort the results on the date
{"$sort": { "_id._id": 1, "Log.date": 1 }},
// Group using $first and $last
{"$group": {
"_id": "$_id",
"firstLog": {"$first": "$Log" },
"lastLog": {"$last": "$Log" }
}}
])
Your mileage may vary, but those approaches may obviate the need to index if this indeed would the the only usage for that index.
I have the following document in the mongodb collection
[{quarter:'Q1',project:'project1',user:'u1',cost:'100'},
{quarter:'Q2',project:'project1',user:'u2',cost:'100'},
{quarter:'Q3',project:'project1',user:'u1',cost:'200'},
{quarter:'Q1',project:'project2',user:'u2',cost:'200'},
{quarter:'Q2',project:'project2',user:'u1',cost:'300'},
{quarter:'Q3',project:'project2',user:'u2',cost:'300'}]
i need to generate an output which will sum the cost based on quarter and project and put it in the format so that it can be rendered in the Extjs chart.
[{quarter:'Q1','project1':100,'project2':200,'project3':300},
{quarter:'Q2','project1':100,'project2':200,'project3':300},
{quarter:'Q3','project1':100,'project2':200,'project3':300}]
i have tried various permutations and combinations of aggregates but couldnt really come up with a pipeline. your help or direction is greatly appreciated
Your cost data appears to be strings, which isn't helping, but assuming you're around that:
The main component is the $cond operator in the document projection, and assuming your data is larger and you want to group the results:
db.mstats.aggregate([
// Optionaly match first depending on what you are doing
// Sum up cost for each quarter and project
{$group: {_id: { quarter: "$quarter", project: "$project" }, cost: {$sum: "$cost" }}},
// Change the "projection" in $group, using $cond to add a key per "project" value
// We use $sum and the false case of 0 to fill in values not in the row.
// These will then group on the key adding the real cost and 0 together.
{$group: {
_id: "$_id.quarter",
project1: {$sum: {$cond:[ {$eq: [ "$_id.project", "project1" ]}, "$cost", 0 ]}},
project2: {$sum: {$cond:[ {$eq: [ "$_id.project", "project2" ]}, "$cost", 0 ]}}
}},
// Change the document to have the "quarter" key
{$project: { _id:0, quarter: "$_id", project1: 1, project2: 1}},
// Optionall sort by quarter
{$sort: {quarter: 1 }}
])
So after doing the initial grouping the document is altered with use of $cond to determine if the value of a key is going to go into the new key name. Essentially this asks if the current value of project is "project1" then put the cost value into this project1 key. And so on.
As we put a 0 value into this new document key when there was no match, we need to group the results again in order to merge the documents. Sorting is optional, but probably what you want for a chart.
Naturally you will have to build this up dynamically and probably query for the project keys that you want. But otherwise this should be what you are looking for.